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Let's try evaluating this limit together. The limit as X goes to infinity of 7 X cubed
plus 5 X plus 3 all over X cubed minus 1. So, this is a rational function and we want
to take the limit as X goes to infinity. So, what would happen if we just tried to use
the limit laws? So, what would I get on top if I just plugged in infinity? So, I've got
7 X cubed plus 5 X plus 3. If I think about taking the limit as X goes to infinity of
that, it's just going to be infinity, because that numerator is just going to get larger
and larger as X gets larger and larger. And same thing with the denominator. If I simply
take the limit of the denominator as X goes to infinity, it's just going to get larger
and larger, because I've got X cubed. So, I see that I have an indeterminate form, infinity
over infinity. This tells me that I need to do some more algebra to determine the answer
to the limit. So, what we're going to do here is a little bit of an algebra trick that's
really useful when taking the limit as X goes to infinity of a rational function. What we're
going to do is divide each term in the numerator and denominator by X to the N where N is the
highest power in the denominator. So, that's a lot of words. Let me show you what I mean.
So, the highest power in the denominator is cubed -- X to the third. So, I'm going to
divide every term by X to the third. That's going to look like this. 7 X over X to the
third plus 5 X over X cubed plus 3 over X cubed on top. And then on the bottom, X cubed
over X cubed minus 1 over X cubed. Now, I can say that these limits are equivalent,
and this is why. So, by dividing every term by X cubed, in a way I sort of have changed
the function because now this function is no longer defined at X equals 0. But, because
I'm looking at the limit as X goes to infinity, I don't really care if the function is or
isn't defined at 0. So, that's why these limits are equivalent and I can use this trick. After
we actually divide every term by whatever -- in this case X cubed, you want to simplify
next. So, I just want to go ahead and simplify each of these terms. In the numerator, 7 X
cubed over X cubed, that's just going to be 7, plus 5 over X squared -- because I have
X over X cubed, so that's going to give me X squared in the denominator -- and then plus
3 over X cubed. And that's all over -- so in my denominator I have X cubed over X cubed,
so that's just going to give me one, minus 1 over X cubed. Having it in this form is
now going to allow me to take the limit pretty easily. Let's just think about it term by
term. Because remember, the limit of the quotient is the quotient of the limits, so I can split
up numerator denominator. And then, since I've just got sums and differences, I can
split it up into each term as well. So, we'll just think about this term by term. So, the
limit as X goes to infinity of 7 is just 7, plus... What about the limit as X goes to
infinity of 5 over X squared? There, my denominator is getting bigger and bigger and bigger, 5
over X squared and X is going to infinity, so my denominator is getting huge and 5 over
a number that's getting bigger and bigger is approaching 0. Think about like 5 over
a trillion, that's a really tiny number. So, this is going to be plus 0. Plus 3 over X
cubed. Same deal. That's getting closer and closer to 0. And then in the denominator,
I have the limit as X goes to infinity of 1 -- that's just 1. Minus 1 over X cubed,
so once again that's going to be something that's approaching zero as that denominator
gets larger and larger. So, if I work this all out, this just becomes 7. So I found that
the limit as X goes to infinity of this function equals 7. So, that tells me I have a horizontal
asymptote on the right hand side and that horizontal asymptote is Y equals 7. Now let
me very quickly also show you how to find the limit as X goes to negative infinity.
In this case it's really not going to be anything very different. Let's just go from this step,
where we've already simplified everything. So, what would change if I plugged in negative
infinity? Well, the 7 is still going to be the same, right? That's just a constant. It's
always 7. So, the limit as X goes to negative infinity of 7 is just 7. What about 5 over
X squared? So, if I plug in negative infinity to the denominator, the denominator is still
getting really big, it's just getting really big in a negative direction. So, you can think
of something like negative a trillion or 5 over negative one billion. So, even though
that number is negative, it's still getting closer and closer to zero. So, all of these
terms that were zero before, are still going to be zero for that reason. The numbers are
negative, but they're still getting closer and closer to 0. So, the limit as X goes to
negative infinity is also 7, so we have the same horizontal asymptote on both sides of
the graph in this case.